5
$\begingroup$

Prove that it is possible to write $\mathbb{R}$ as a union $\mathbb{R} = \bigcup_{i\in I} A_{i}$ where $A_{i} \cap A_{j} = \phi$ if $i\neq j$, $i, j \in I$, and such that each $A_{i}$ and $I$ are uncountable sets.

It's fairly straight forward to find a union where either the $A_{i}$'s are countably infinite and $I$ is uncountably infinite or $I$ is countably infinite and the $A_{i}$'s are uncountably infinite.

One idea I had to get them to be both uncountable was to break the real line up into intervals $[i, i+\Delta i)$ and take the limit as $\Delta i$ goes to zero. But I don't think this works because either $\Delta i$ is zero or it's greater than zero. If it's zero then $I$ is uncountable and $A$ only contains one real number, and is therefore countable. If $\Delta i$ is greater than zero, then it's possible to enumerate the intervals, so $I$ is countable.

Another idea was to take the set of binary strings of infinite length and try to find a way to map each string to a uncountably infinite set of other binary strings of infinite length. I thought maybe this could be done by splitting up the strings into sets ($A_{i}$) depending on their first n digits, and then taking the limit as n goes to infinity.

The problem with this is that if we give infinite digits then we've specified the string itself and the set $A_{i}$ can only contain one string. Otherwise the first n digits can be mapped to a natural number and so $I$ is countable. Same problem.

Any help with this is appreciated.

Thanks

$\endgroup$
  • 1
    $\begingroup$ The idea of taking a limit of a finite or countable number of sets will not work, for the reasons you mentioned. What if you decompose the reals by looking at their odd-digits? $\endgroup$ – Michael Mar 4 '17 at 23:03
  • $\begingroup$ It is tantamount to finding a surjective (no need for an actual bijection $\Bbb R\to\Bbb R^2$) function $f:\Bbb R\to\Bbb R$ such that $\operatorname{card}(f^{-1}(y))=2^{\aleph_0}$ for all $y$. This can be made "rather explicitly" using Hilbert-Peano curves (and perhaps some other method I am not fully aware of). See, for instance, my answer here. $\endgroup$ – user228113 Mar 4 '17 at 23:19
  • $\begingroup$ This method has the consequence of showing that you can also require the $A_i$s to be all closed subsets. $\endgroup$ – user228113 Mar 4 '17 at 23:28
  • $\begingroup$ You may find this interesting. $\endgroup$ – goblin Mar 5 '17 at 3:04
6
$\begingroup$

A proof (which, as yo' has pointed out, can be as contrustive as you like):

Since $|\mathbb{R}| = |\mathbb{R} \times \mathbb{R}| $, there exists a bijection $f$ from $\mathbb{R} \to \mathbb{R} \times \mathbb{R} $. Then $\mathbb{R} = \bigcup_{a \in \mathbb{R}} f^{-1}(\mathbb{R},a)$ where $f^{-1}(\mathbb{R},a) = \{b \in \mathbb{R}: f(b) = (c,a)$ for some $c \in \mathbb{R} \}$.

$\endgroup$
  • $\begingroup$ This can be very constructive: mathoverflow.net/questions/126069/… $\endgroup$ – yo' Mar 4 '17 at 23:19
  • $\begingroup$ @yo': That's really neat! $\endgroup$ – Kyle Mar 4 '17 at 23:22
  • $\begingroup$ @KyleGannon People say that each $f^{-1}(\mathbb{R},a)$ is the preimage of one veritical line in the plane. But it seems to me that each $f^{-1}(\mathbb{R},a)$ should be the preimage of one horizontal line in the plane. For example, $f^{-1}(\mathbb{R},\sqrt 2)$ is the set $\{b \in \mathbb{R}: f(b) = (c,\sqrt2)$ for some $c \in \mathbb{R} \}$. Is my understanding correct? $\endgroup$ – PropositionX Mar 13 '17 at 1:47
4
$\begingroup$

Hint: $\Bbb{R}^2 = \bigcup_{x\in \Bbb{R}} (\{x\} \times \Bbb{R)}$.

$\endgroup$
0
$\begingroup$

Take power set of whole numbers minus the singleton sets. and real numbers formed using the numbers of an element of that power set. Power set is also uncountable and numbers formed using an element of power set minus singleton set is also uncountable.

$\endgroup$
0
$\begingroup$

Here is an explicit construction, as mentioned in @Michael's comment: Let each $ A_i $ be the set of real numbers whose odd-numbered digits correspond to some real number $ x_i $. The set $ A $ then has a trivial bijection with $\mathbb {R} $ , as does each $ A_i $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.